Cho a,b,c,>0 sao cho $\frac{1}{a}$+$\frac{1}{b}$+ $\frac{1}{c}$=4 CMR : P= $\frac{1}{2a+b+c}$+$\frac{1}{2b+a+c}$+ $\frac{1}{2c+a+b}$$\leq$ 2

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1. Đáp án:

   áp dụng bđt $\frac{1}{a}+\frac{1}{b}≥\frac{4}{a+b}$

   dấu = xảy ra khi a=b

   Giải thích các bước giải:

   ta có

   $\frac{1}{2a+b+c}=\frac{1}{(a+b)+(a+c)} ≤\frac{1}{4}(\frac{1}{a+b}+\frac{1}{b+c}) ≤ \frac{1}{4}(\frac{1}{4}(\frac{1}{a}+\frac{1}{b})+\frac{1}{4}(\frac{1}{a}+\frac{1}{c}))=\frac{1}{16}(\frac{2}{a}+\frac{1}{b}+\frac{1}{c})$

   tương tự ta có:

   $\frac{1}{a+2b+c} ≤ \frac{1}{16}(\frac{1}{a}+\frac{2}{b}+\frac{1}{c})$

   $\frac{1}{a+b+2c} ≤ \frac{1}{16}(\frac{1}{a}+\frac{1}{b}+\frac{2}{c})$

   suy ra $P ≤\frac{1}{16}4(\frac{1}{a}+\frac{1}{b}+\frac{1}{c})=1$

   dấu = xảy ra khi $a=b=c=\frac{3}{4}$

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